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Let $a$ and $b$ be positive real numbers such that $a + b = 1$. Prove that $$a^a\cdot b^b+a^b \cdot b^a≤1$$

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  • $\begingroup$ For $a=b=0.5$ it doesn't appear to be true. Typo somewhere? $\endgroup$
    – Joffan
    May 30, 2016 at 7:28
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    $\begingroup$ Instead of $a^b+b^a$, it should be $a^b\cdot b^a$. $\endgroup$
    – Roby5
    May 30, 2016 at 7:33
  • $\begingroup$ @Roby5. You are right. $\endgroup$ May 30, 2016 at 7:36

2 Answers 2

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By the Jensen's inequality, $$ \log[a^ab^b]=a\log a+b\log b\leq\log[a^2+b^2]\implies a^ab^b\leq a^2+b^2. $$ Similarly, $$ \log[a^bb^a]=a\log b+b\log a\leq \log[ab+ba]\implies a^bb^a\leq 2ab. $$ Summing then gives: $$ a^ab^b+a^bb^a\leq a^2+b^2+2ab=(a+b)^2=1. $$

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Hint: Since $a + b = 1$, we can use the Weighted Arithmetic Mean - Geometric Mean inequality to get $$\sqrt[a+b]{a^a b^b} \leq a^2 + b^2.$$ Similarly, write the inequality for the other term to get the desired result.

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